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To find the limit $$\lim_{n \rightarrow \infty} \sqrt{n} \int_0^1 \left(\frac{\sin t}{t}\right)^n dt,$$ I attempted to use Laplace's method. We can express the given integral as $$\int_0^1 \exp \left(n \ln \left(\frac{\sin x}{x}\right)\right) dx$$ However, I encountered an issue with this approach. In Laplace's method, we require the function $f(x)=\ln \left(\frac{\sin x}{x}\right)$ to be twice differentiable. Additionally, the global maximum of $f(x)$ within the integration range must be unique and not located at the boundary points. Unfortunately, in this case, the maximum occurs at the boundary point $0$, which prevents me from applying Laplace's method as intended.

I have been thinking about this problem for quite some time, but I haven't been able to come up with any promising ideas. I would really appreciate any guidance or suggestions. Thank you in advance.

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4 Answers 4

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You can expand the interval of integration to $[-1,1]$ since $t\mapsto \frac{\sin t}{t}$ is an even positive function on that interval. Then you can apply Laplace's method.

A more direct approach is to use a change of variables and dominated convergence:

Set $I_n=\sqrt{n}\int^1_0\Big(\frac{\sin t}{t}\Big)^n\,dt$. On $[0,1]$ we have that

$$0<1-\frac{t^2}{6}\leq \frac{\sin t}{t}\leq 1-\frac{t^2}{6}+\frac{t^4}{120}$$

The change of variables $u=\sqrt{n} t$ yields $$\int^{\sqrt{n}}_0\Big(1-\frac{u^2}{6n}\Big)^n\,du\leq I_n\leq \int^{\sqrt{n}}_0\Big(1-\frac{u^2}{6n}+\frac{u^4}{120n^2}\Big)^n\,du\leq\int^{\sqrt{n}}_0e^{-\Big(\frac{u^2}{6}-\frac{u^4}{120n}\Big)}\,du$$

  • Notice that $\big(1-\frac{u^2}{6n}\Big)^n\leq e^{-\frac{u^2}{6}}$.

  • Notice that on $[0,\sqrt{n}]$, $\frac{u^4}{120 n}\leq \frac{u^2}{120}$ and so, $$\frac{u^2}{6}-\frac{u^4}{120n}\geq\frac{19u^2}{120}$$

This means that $$e^{-\Big(\frac{u^2}{6}-\frac{u^4}{120n}\Big)} \leq e^{-\frac{19u^2}{120}}, \qquad 0\leq u\leq \sqrt{n}$$

Since $f_a(u)=e^{-au^2}\in L_1(\mathbb{R})$ for all $a>0$, an application of dominated convergence gives

$$\lim_nI_n=\int^\infty_0e^{-u^2/6}\,du$$

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  • $\begingroup$ Good answer +1! Using the Laplace method is equivalent to your answer, the proof of Laplace method uses the same idea as yours. $\endgroup$
    – Kroki
    Commented May 15, 2023 at 3:50
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    $\begingroup$ $$\int_{0}^{\infty} e^{-u^2/6} \,du = \frac{\sqrt{6 \pi}}{2}$$ $\endgroup$
    – Bumblebee
    Commented May 15, 2023 at 4:16
  • $\begingroup$ @Oliver Díaz I would like to ask a possibly silly question. I understand the conditions for dominated convergence are satisfied, but I don't understand how to apply them to obtain $\lim _n I_n=\int_0^{\infty} e^{-u^2 / 6} d u$? $\endgroup$
    – unicornki
    Commented May 15, 2023 at 4:37
  • $\begingroup$ Using the dominated convergence, you can interchange the integral and limit as $n\to \infty$ $\endgroup$
    – Kroki
    Commented May 15, 2023 at 4:41
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    $\begingroup$ I got it, thank you very much. $\endgroup$
    – unicornki
    Commented May 15, 2023 at 4:51
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Hint: $$\int_{0}^1 \left(\frac{\sin t}{t}\right)^n \mathrm d t =\frac12\int_{-1}^1 \left(\frac{\sin t}{t}\right)^n\mathrm d t$$

Can you apply Laplace method now?

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Remarks: Here, we obtain the lower and upper bounds of $I_n$ which do not involve $n$ in the integrands (see (6)). In contrast, if the integrand involves $n$ i.e. $f_n(x)$, perhaps we need some convergence theorem to ensure that $\lim_{n\to \infty} \int_0^{\sqrt n} f_n(x)\,\mathrm{d} x = \int_0^\infty \lim_{n\to \infty} f_n(x) \,\mathrm{d}x$.

Let $$I_n := \sqrt{n} \int_0^1 \left(\frac{\sin t}{t}\right)^n \,\mathrm{d} t.$$

Using the known inequality $\sin t \le t - t^3/6 + t^5/120$ for all $t \ge 0$, we have, for all $t \in (0, 1]$, $$\frac{\sin t}{t} \le 1 - t^2/6 + t^4/120 \le \mathrm{e}^{-t^2/6}. \tag{1}$$ (Proof of the right inequality: Let $f(t) := -t^2/6 - \ln (1 - t^2/6 + t^4/120)$. We have $f'(t) = \frac{t^3(8 - t^2)}{3t^4 - 60t^2 + 360} \ge 0$ on $t\in (0, 1]$. Also, $f(0) = 0$. Thus, $f(t)\ge 0$ on $(0, 1]$.)

Using (1), we have $$I_n \le \sqrt{n} \int_0^1 \mathrm{e}^{-nt^2/6} \,\mathrm{d} t = \int_0^{\sqrt n} \mathrm{e}^{-u^2/6}\,\mathrm{d} u. \tag{2}$$

On the other hand, using the known inequality $\sin t \ge t - t^3/6$ for all $t \ge 0$, noting that $1 - t^2/6 \ge 0$ on $(0, 1]$, we have $$I_n \ge \sqrt{n}\int_0^1 (1 - t^2/6)^n\,\mathrm{d}t = \int_0^{\sqrt n} \left(1 - \frac{u^2}{6n}\right)^n\,\mathrm{d} u. \tag{3}$$

We have, for all $u \in [0, \sqrt n]$, $$\left(1 - \frac{u^2}{6n}\right)^n \ge \mathrm{e}^{-u^2/6}\left(1 - \frac{u^4}{n}\right). \tag{4}$$ (The proof is given at the end.)

Using (3) and (4), we have $$I_n \ge \int_0^{\sqrt n} \mathrm{e}^{-u^2/6}\left(1 - \frac{u^4}{n}\right)\,\mathrm{d} u = \int_0^{\sqrt n} \mathrm{e}^{-u^2/6}\,\mathrm{d} u - \frac{1}{n} \int_0^{\sqrt n} \mathrm{e}^{-u^2/6}u^4\,\mathrm{d} u.\tag{5}$$

Using (2) and (5), we have $$\int_0^{\sqrt n} \mathrm{e}^{-u^2/6}\,\mathrm{d} u - \frac{1}{n} \int_0^{\sqrt n} \mathrm{e}^{-u^2/6}u^4\,\mathrm{d} u \le I_n \le \int_0^{\sqrt n} \mathrm{e}^{-u^2/6}\,\mathrm{d} u. \tag{6}$$

By the squeeze theorem, we have $$\lim_{n\to \infty} I_n = \frac12\sqrt{6\pi}$$ where we use $$\int_0^\infty \mathrm{e}^{-u^2/6}\,\mathrm{d} u = \frac12\sqrt{6\pi}, \quad \int_0^{\infty} \mathrm{e}^{-u^2/6}u^4\,\mathrm{d} u = \frac{27}{2}\sqrt{6\pi}.$$

We are done.


Proof of (4):

We only need to prove the case that $1 - u^4/n > 0$ i.e. $u < \sqrt[4]{n}$.

Taking logarithm, it suffices to prove that $$F(u) := n\ln \left(1 - \frac{u^2}{6n}\right) + \frac{u^2}{6} - \ln\left(1 - \frac{u^4}{n}\right).$$

We have, for all $u\in [0, \sqrt[4]{n})$, $$F'(u) = \frac{u^3(u^4 - 12u^2 + 71n)}{3(6n - u^2)(n - u^4)} > 0.$$ Also, we have $F(0) = 0$. Thus, $F(u) \ge 0$ on $[0, \sqrt[4]{n})$.

We are done.

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For the fun of going beyond the limit

Consider $$ \log \left(\frac{\sin (t)}{t}\right)=-\sum_{k=1}^\infty \frac{4^{k-1}\, \left| B_{2 k}\right| }{k^2\,\Gamma (2 k)}\, t^{2k}$$

Truncate to some order, isolate the first term, exponentiate at using Taylor series

So $$\left(\frac{\sin (t)}{t}\right)^n=e^{-\frac{n }{6}t^2}\left( 1+n\sum_{k=2}^\infty \alpha_k\, t^{2k}\right)$$ where the first coefficients are $$\left( \begin{array}{cc} k & \alpha_k \\ 2 & -\frac{1}{180} \\ 3 & -\frac{1}{2835} \\ 4 & \frac{7 n-12}{453600} \\ 5 & \frac{11 n-12}{5613300} \\ 6 & -\frac{21021 n^2-153868 n+132672}{735566832000} \\ \end{array} \right)$$

So, for the integral $$\sqrt{\frac{3 \pi }{2 n}}\text{erf}\left(\sqrt{\frac{n}{6}}\right)+\sqrt{\frac{3n}{2}}\sum_{k=2}^\infty \alpha_k\,\left(\frac{6}{n}\right)^k \left(\Gamma \left(\frac{2k+1}{2}\right)-\Gamma \left(\frac{2k+1}{2},\frac{n}{6}\right)\right)$$ Using the only coefficients given in the table, this gives pretty decent results even for small values of $n$ $$\left( \begin{array}{ccc} n & \text{approximation} & \text{"exact" value}\\ 1 & 0.94608307 & 0.94608307 \\ 2 & 0.89733956 & 0.89733956 \\ 3 & 0.85318077 & 0.85318077 \\ 4 & 0.81309084 & 0.81309084 \\ 5 & 0.77661723 & 0.77661723 \\ 6 & 0.74336258 & 0.74336256 \\ 7 & 0.71297757 & 0.71297754 \\ 8 & 0.68515479 & 0.68515476 \\ 9 & 0.65962342 & 0.65962338 \\ 10 & 0.63614458 & 0.63614453 \\ \end{array} \right)$$

Using asymptotics $$ \int_0^1 \left(\frac{\sin (t)}{t}\right)^n\, dt=\sqrt{\frac{3 \pi }{2 n}}\left( 1-\frac{3}{20 n}-\frac{13}{1120 n^2}+O\left(\frac{1}{n^3}\right) \right)$$

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  • $\begingroup$ Related: math.stackexchange.com/q/4638733 $\endgroup$
    – Gary
    Commented May 15, 2023 at 11:20
  • $\begingroup$ @Gary. Thanks for the link ! I was about to ping you because of the upper limit equal to $1$. Don't you think that the asymptotic needs a bit more ? Cheers :-) $\endgroup$ Commented May 15, 2023 at 11:51
  • $\begingroup$ Any positive upper limit of integration will produce the same asymptotic expansion, provided that one seeks an asymptotic expansion in terms of elementary functions and in the sense of Poincaré. $\endgroup$
    – Gary
    Commented May 15, 2023 at 11:58

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